Log2(x-2)+log2(x-3)=1

To solve the given logarithmic equation:

\( \log_2(x-2) + \log_2(x-3) = 1 \)

We can use the product rule of logarithms, which states that \( \log_a(b) + \log_a(c) = \log_a(bc) \). Applying this rule to the equation:

\( \log_2((x-2)(x-3)) = 1 \)

This simplifies to:

\( \log_2(x^2 - 5x + 6) = 1 \)

Now, we can rewrite the equation in exponential form:

\( 2^1 = x^2 - 5x + 6 \)

Simplify this further:

\( 2 = x^2 - 5x + 6 \)

Rearrange the equation to set it equal to zero:

\( x^2 - 5x + 4 = 0 \)

Now, factor the quadratic equation:

\( (x-4)(x-1) = 0 \)

Set each factor to zero to find the possible solutions:

\( x-4 = 0 \) or \( x-1 = 0 \)

\( x = 4 \) or \( x = 1 \)

Therefore, the solutions to the equation are \( x = 4 \) and \( x = 1 \).